On the geometry of full points of abstract unitals

TL;DR

This paper explores the detailed combinatorial and geometric properties of full points in finite order abstract unitals, enhancing understanding of their structure for projective embedding studies.

Contribution

It provides a detailed description of the combinatorial and geometric structure of full points in finite order abstract unitals, expanding on previous foundational work.

Findings

Characterization of full points in finite order abstract unitals

Insights into the geometric structure of full points

Implications for projective embeddings of abstract unitals

Abstract

The concept of full points of abstract unitals has been introduced by Korchm\'aros, Siciliano and Sz\H{o}nyi as a tool for the study of projective embeddings of abstract unitals. In this paper we give a more detailed description of the combinatorial and geometric structure of the sets of full points in abstract unitals of finite order.

Tables

Table 1. Table 1. BBT unitals of order 3 3 3

Full points	Group of perspectivities	Unitals
2	$C_2$	477
2	$C_3$	94
2	$C_4$	290

Table 2. Table 2. KRC unitals of order 3 3 3

Full points	Group of perspectivities	Unitals
2	$C_2$	1015
2	$C_3$	379
2	$C_4$	897
3	$S_4$	6

Table 3. Table 3. KNP unitals of order 4 4 4

Full points	Group of perspectivities	Unitals
2	$C_2$	93
2	$C_4$	71
2	$C_5$	107
2	$C_6$	5
3	$A_5$	2
3	$C_2\times C_2$	1
3	$C_4$	32
3	$C_5$	30
3	$S_5$	3
4	$C_5$	8
5	$C_5$	165
6	$C_5⋊C_4$	72
6	$D_{10}$	53

Table 4. Table 4. KNP unitals with large full point sets

set	property	cardinality
$\Omega$	at least one large full point set	206
$A$	all large full point sets form a block	74
$B$	all large full point sets are contained in a block	80
$\overline{B}$	some large full point sets are not contained a block	126
$C$	no large full point set is contained in a block	1

Table 5. Table 5. Full point regularity

Library	Unitals	FPR	SFPR
BBT	909	815	815
KRC	4466	4081	4081
KNP	1777	1586	1582

Equations

$$|F_U(b_1,b_2)|\le \{\begin{array}{ll}{n}^2-n& \text{if b1,b2 have a point in common,}\\ n^2-1& \text{if b1,b2 are disjoint.}\end{array}$$

$$|S_P|=\{\begin{array}{ll}{n}^2-n& \text{if b1,b2 have a point in common,}\\ n^2-1& \text{if b1,b2 are disjoint.}\end{array}$$

$${\mathrm{Persp}}_{b_2}(b_1)=⟨{\pi }_{b_1,P,b_2}{\pi }_{b_2,Q,b_1}\mid P,Q\in F_U(b_1,b_2)⟩.$$

$$k+{\left(n+1\right)}^2+k\left(n+1\right)\left(n^2-n-2\right)$$

$$b_s=\{P_{s,1},\dots ,P_{s,n+1}\},s=1,2,3.$$

$$A_i=(0,1,{\epsilon }^i),B_j=({\epsilon }^j,0,1),C_k=(1,{\epsilon }^k,0),$$

Full text

Support provided from the National Research, Development and Innovation Fund of Hungary, financed under the 2018-1.2.1-NKP funding scheme, within the SETIT Project 2018-1.2.1-NKP-2018-00004. Partially supported by OTKA grants 119687 and 115288.

1. Introduction

An abstract unital of order $n$ is a $2$-$(n^{3}+1,n+1,1)$ design. We say that an abstract unital $(X,B)$ is embedded in a projective plane $\Pi$ if $X$ consists of points of $\Pi$ and each block $b\in B$ has the form $X\cap\ell$ for some line $\ell$ of $\Pi$. For results on projective embeddings of abstract unitals see [12] and the references therein.

Let $U=(X,B)$ be an abstract unital of order $n$ and fix two blocks $b_{1},b_{2}$. Using the terminology of [12], we say that $P$ is a full point with respect to $(b_{1},b_{2})$ if $P\not\in b_{1}\cup b_{2}$ and for each $Q\in b_{1}$, the block connecting $P$ and $Q$ intersects $b_{2}$. In other words, there is a well defined projection $\pi_{b_{1},P,b_{2}}$ from $b_{1}$ to $b_{2}$ with center $P$. We denote by $F_{U}(b_{1},b_{2})$ the set of full points of $U$ w.r.t. the blocks $b_{1},b_{2}$. Clearly, $F_{U}(b_{1},b_{2})=F_{U}(b_{2},b_{1})$.

The structure of the paper is as follows. The main result of this paper is proved in Section 3. It shows that for any abstract unital of order $q$, which is projectively embedded in the Galois plane $\mathsf{PG}(2,q^{2})$, the set of full points of two disjoint blocks are contained in a line. Moreover, the perspectivities of two disjoint blocks generate a semi-regular cyclic permutation group. In Section 4, we extend the results of [12] by giving a complete description on the structure of full points in the classical Hermitian unitals. Section 5 gives an overview of computational results about full points in abstract unitals of order $3$ and $4$, which belong to known classes [1, 6, 14, 13]. For the computation we developed and used the GAP package UnitalSZ [16].

2. Combinatorial properties of the set of full points

2.1. Bounds on the number of full points

We start with an easy observation on the number of full points of two blocks $b_{1},b_{2}$ of $U$. The result seems to be rather weak.

Lemma 2.1**.**

Let $U=(X,B)$ be an abstract unital of order $n\geq 2$. Then

[TABLE]

Proof.

For a fixed point $P\in b_{1}$ we define the set $S^{\prime}_{P}$ as the union of the blocks connecting $P$ with $Q\in b_{2}\setminus b_{1}$, and the set $S_{P}=S^{\prime}_{P}\setminus(b_{1}\cup b_{2})$. Clearly,

[TABLE]

As $F_{U}(b_{1},b_{2})\subseteq S_{P}$, the lemma follows. ∎

In most (but not all) known examples of abstract unitals, the set of full points is contained in a block. This motivates the following definition.

Definition 2.2**.**

Let $U=(X,B)$ be an abstract unital and $b_{1},b_{2}\in B$ disjoint blocks.

1. (i)

The triple $\left(U,b_{1},b_{2}\right)$ is full point regular if the set of full points $F_{U}{\left(b_{1},b_{2}\right)}\subseteq c$ for some block $c\in B$ such that $b_{1}\cap c=b_{2}\cap c=\emptyset$. 2. (ii)

The abstract unital $U$ is full point regular if for any two disjoint blocks $b_{1},b_{2}$ the triple $\left(U,b_{1},b_{2}\right)$ is full point regular.

2.2. Full points and perspectivities

By definition, any full point $P$ of the blocks $b_{1},b_{2}$ defines a bijective map $\pi_{b_{1},P,b_{2}}:b_{1}\to b_{2}$; we call it the perspectivity with center $P$.

Definition 2.3**.**

Let $b_{1},b_{2}$ be blocks of the abstract unital $U$. Define the group of perspectivities of $b_{1}$ as

[TABLE]

It is easy to see that $\operatorname{Persp}_{b_{2}}(b_{1})$ and $\operatorname{Persp}_{b_{1}}(b_{2})$ are isomorphic permutation groups, the former acting on $b_{1}$ and the latter acting on $b_{2}$. For different full points $Q,R$, the perspectivities $\pi_{b_{1},Q,b_{2}}$ and $\pi_{b_{1},R,b_{2}}$ are different. This implies $|\operatorname{Persp}_{b_{2}}(b_{1})|\geq|F_{U}(b_{1},b_{2})|$. In particular, $\operatorname{Persp}_{b_{2}}(b_{1})$ is nontrivial if $|F_{U}(b_{1},b_{2})|>1$. An important case will be when $\operatorname{Persp}_{b_{2}}(b_{1})$ is a cyclic semi-regular permutation group on $b_{1}$.

2.3. Dual $k$-nets in abstract unitals

We will present examples of abstract unitals when the set of full points w.r.t. the blocks $b_{1},b_{2}$ form a third block $b_{3}$. More generally, we introduce the concept of an embedded dual $k$-net of an abstract unital. An abstract $k$-net is a structure consisting of a set $X$ of points and a set $B$ of blocks, which is partitioned into $k$ disjoint families $B_{1},\ldots,B_{k}$ for which the following hold: (1) every point is incident with exactly one block of every $B_{i}$, $(i=1,\ldots,k)$; (2) two lines of different families have exactly one point in common; (3) there exist $3$ lines belonging to $3$ different $B_{i}$, and which are not incident with the same point. See [3, 5] as reference on abstract $k$-nets.

Definition 2.4**.**

Let $U=(X,B)$ be an abstract unital of order $n$ and $k\geq 3$ and integer. We say that the blocks $b_{1},\ldots,b_{k}$ form an embedded dual $k$-net in $U$, if the following hold for all $1\leq i<j\leq k$:

1. (i)

$b_{i}\cap b_{j}=\emptyset$. 2. (ii)

For all $P\in b_{i}$, $Q\in b_{j}$, the block containing $P,Q$ intersects all $b_{1},\ldots,b_{k}$ in a point.

It is clear that for an embedded dual $k$-net $b_{1},\ldots,b_{k}$ of $U$, $b_{3}\cup\cdots\cup b_{k}\subseteq F_{U}(b_{1},b_{2})$. The converse needs some explanation.

Lemma 2.5**.**

Let $U$ be an abstract unital of order $n$, $k\geq 3$ an integer and $b_{1},\ldots,b_{k}$ blocks of $U$.

1. (i)

If $b_{3}\subseteq F_{U}(b_{1},b_{2})$, then $b_{1}$ and $b_{2}$ are disjoint. 2. (ii)

If $b_{3}\subseteq F_{U}(b_{1},b_{2})$, then $b_{1}\subseteq F_{U}(b_{2},b_{3})$ and $b_{2}\subseteq F_{U}(b_{1},b_{3})$. 3. (iii)

If $b_{3}\cup b_{4}\subseteq F_{U}(b_{1},b_{2})$, then $b_{3}$ and $b_{4}$ are disjoint. 4. (iv)

The blocks $b_{1},\ldots,b_{k}$ form an embedded dual $k$-net if and only if $b_{3}\cup\cdots\cup b_{k}\subseteq F_{U}(b_{1},b_{2})$.

Proof.

(i) Assume that $\{Z\}=b_{1}\cap b_{2}$ and $b_{3}\subseteq F_{U}(b_{1},b_{2})$. Clearly, $b_{3}$ is disjoint from $b_{1}\cup b_{2}$. Fix an arbitrary point $P\in b_{1}\setminus\{Z\}$. Each point $R\in b_{3}$ projects $P$ to $b_{2}\setminus\{Z\}$. Hence, there are points $R_{1},R_{2}\in b_{3}$ such that $\pi_{b_{1},R_{1},b_{2}}(P)=\pi_{b_{1},R_{2},b_{2}}(P)$. This means that $P\in b_{3}$, a contradiction. (ii) For any $P_{1}\in b_{1}$, $P_{3}\in b_{3}$, the block $P_{1}P_{3}$ intesects $b_{2}$. Now fix $P_{1}$ and let $P_{3}$ run through $b_{3}$ in order to obtain the bijection $\pi_{b_{3},P_{1},b_{2}}$. Thus, $P_{1}\in F_{U}(b_{2},b_{3})$. Since this holds for all $P_{1}\in b_{1}$, the claim follows. For (iii) it suffices to show $b_{1}\subseteq F_{U}(b_{3},b_{4})$. Take $P\in b_{1}$, $Q\in b_{3}$ arbitrary points. From $Q$, $P$ projects to $R\in b_{2}$ and using $b_{2}\subseteq F_{U}(b_{1},b_{4})$, $P$ projects to $S\in b_{4}$ from $R$. Hence, $Q$ projects to $b_{4}$ from $P$.

The “only if” part of (iv) follows from the definition. Assume now $b_{3}\cup\cdots\cup b_{k}\subseteq F_{U}(b_{1},b_{2})$. By (i) and (iii), all blocks $b_{1},\ldots,b_{k}$ are disjoint. For the indices $3\leq i<j\leq k$, there is an injective map $\alpha\mid b_{1}\times b_{2}\rightarrow b_{i}\times b_{j}$ mapping $\left(P_{1},P_{2}\right)\mapsto\left(P_{i},P_{j}\right)$ with collinear quadruple $P_{1},P_{2},P_{i},P_{j}$. Moreover $\alpha$ is bijective, hence any pair of points $\left(P_{i},P_{j}\right)\in b_{i}\times b_{j}$ determines a block $b^{\prime}$ of $U$ such that $b^{\prime}\cap b_{i}=P_{i}$, $i=1,2$. The block joining $P_{1}$ and $P_{2}$ intersects any block $b_{s}\subseteq F_{U}{\left(b_{1},b_{2}\right)}$ in $P_{s}$ for $3\leq s\leq k$, therefore $b_{1},\ldots,b_{k}$ form a dual $k$-net in $U$.∎

2.4. Bounds on dual $k$-nets in abstract unitals

For embedded dual $k$-nets, the trivial bound is $k\leq n+1$. With some elementary counting, we can improve this to $k\leq n-1$. This implies that an abstract unital of order $3$ has no embedded dual $3$-nets.

Proposition 2.6**.**

Let $U$ be an abstract unital of order $n\geq 3$.

1. (i)

If $U$ has an embedded dual $k$-net $\{b_{1},\ldots,b_{k}\}$, then $k\leq n-1$. 2. (ii)

For two blocks $b_{1},b_{2}$, $F_{U}(b_{1},b_{2})$ cannot contain more than $n-3$ blocks.

Proof.

(i) Let us assume that $k>n-1$ and let $\mathcal{P}=b_{1}\cup b_{2}\cup\ldots\cup b_{k}$. Any block of $U$ intersects $\mathcal{P}$ in [math], $1$, $k$ or $n+1$ points, the latter being the blocks $b_{i}$ themselves. W.l.o.g. consider the disjoint blocks $b_{1},b_{2}$. Any pair of points chosen from $b_{1}$ and $b_{2}$ determines the unique block in $B$ which is a $k$-secant to $\mathcal{P}$, therefore the number of $k$-secants is $\left(n+1\right)^{2}$. Then, fix an arbitrary block $b_{i}$ of the dual $k$-net and a point $P$ on the block $b_{i}$. The number of $1$-secant blocks on $P$ is $n^{2}-n-2$. Thus the number $1$-secant blocks to $\mathcal{P}$ is $k\left(n+1\right)\left(n^{2}-n-2\right)$. Since $\left|B\right|=n^{2}\left(n^{2}-n+1\right)$ we have

[TABLE]

which gives $n^{3}-3n^{2}+n+1\leq 0$ by $k\geq n\geq 3$, a contradiction.

(ii) If $F_{U}(b_{1},b_{2})$ contains the $k-2$ blocks $b_{3},\ldots,b_{k}$, then $\{b_{1},\ldots,b_{k}\}$ is an embedded dual $k$-net in $U$ by Lemma 2.5(iv). Hence, $k-2\leq n-3$ by (i). ∎

2.5. Embedded dual $3$-nets and latin squares

An embedded dual $3$-net $\{b_{1},b_{2},b_{3}\}$ determines a latin square $L$ of order $n+1$ in the following way. Label the points of $b_{1},b_{2},b_{3}$ by the set $\{1,\ldots,n+1\}$:

[TABLE]

For $i,j\in\{1,\ldots,n+1\}$, let $c$ be the block connecting $P_{1,i}$ and $P_{2,j}$. Define $s$ by $\{P_{3,s}\}=b_{3}\cap c$ and write $s$ in row $i$ and column $j$ of $L$. Choosing a different labeling for $b_{1},b_{2},b_{3}$ results in an isotope latin square. By reordering the three blocks, one gets conjugate or parastrophe latin squares. The set of all parastrophes of a latin square $L$ is also called the main class of $L$. Latin squares are naturally related to (the multiplication tables of) finite quasigroups. See [9, Section 1.4] for more details and further references on conjugacy and parastrophy of latin squares.

A property which, for each class $C$, either holds for all members of $C$ or for no member of $C$ is said to be a class invariant. Properties of the underlying (dual) $3$-nets are main class invariants of the corresponding coordinate latin square. In particular, the groups of perspectivities can be defined for (dual) $3$-nets and they are useful examples of main class invariants. In the primal setting, perspectivities of 3-nets have been presented in [3] and [5].

Let $L$ be a latin square of order $n$. We say that $L$ is group-based if it is a parastrophe to the Cayley table of a group $G$ of order $n$. As the group $G$ only depends on the main class of $L$, the following concept is well-defined.

Definition 2.7**.**

Let $\mathcal{B}=\{b_{1},b_{2},b_{3}\}$ be an embedded dual $3$-net of the abstract unital $U$. We say that $\mathcal{B}$ is cyclic, if the corresponding latin square is a parastrophe of the Cayley table of the cyclic group of order $n+1$, where $n$ is the order of $U$.

Proposition 2.8**.**

Let $U$ be an abstract unital of order $n$ and $\mathcal{B}=\{b_{1},b_{2},b_{3}\}$ be an embedded dual $3$-net of $U$. The following are equivalent:

1. (i)

$\mathcal{B}$* is cyclic.* 2. (ii)

$\operatorname{Persp}_{b_{i}}(b_{j})$* is the cyclic group of order $n+1$ for all $1\leq i,j\leq 3$, $i\neq j$.*

Proof.

Let $L$ be the latin square associated to $\mathcal{B}$. By [3, Proposition 1.2], (ii) implies that the rows of $L$ are elements of the cyclic group of order $n$, hence $L$ is cyclic and (i) holds. Conversely, assume that $\mathcal{B}$ is labeled in such a way that the the coordinate latin square $L$ is the Cayley table of the cyclic group. Then [3, Theorem 6.1] implies (ii). ∎

3. Full point regularity of embedded unitals

The questions on the embeddings of abstract unitals in projective planes are long studied problems, with special focus on the embeddings of abstract unitals of order $q$ in the desarguesian plane $\mathsf{PG}(2,q^{2})$. Korchmáros, Siciliano and Szőnyi [12] developed the method of full points for the embedding problem. The main tool is the group of perspectivities of unital blocks. We notice that while the permutation group $\operatorname{Persp}_{b_{2}}(b_{1})$ depends only on the abstract unital structure of $U=(X,B)$, we may be able compute it more easily when a projective embedding of $U$ is given.

Although the definition of the group of perspectivities works for intersecting blocks $b_{1},b_{2}$, in the sequel, we will only deal with the case when $b_{1},b_{2}$ are disjoint. The next definition gives a stronger version of the full point regular property, using the structure of the group of perspectivities.

Definition 3.1**.**

Let $U=(X,B)$ be an abstract unital and $b_{1},b_{2}\in B$ disjoint blocks.

1. (i)

If $\left(U,b_{1},b_{2}\right)$ is a full point regular triple and $\operatorname{Persp}_{b_{2}}(b_{1})$ is a cyclic semi-regular permutation group of $b_{1}$, then $\left(U,b_{1},b_{2}\right)$ is said to be strongly full point regular. 2. (ii)

The abstract unital $U$ is strongly full point regular if for any two disjoint blocks $b_{1},b_{2}$ the triple $\left(U,b_{1},b_{2}\right)$ is strongly full point regular.

Notice that $U$ is strongly full point regular if it has no full points at all. The next two lemmas deal with elementary properties of the groups of affinities of projective lines in $\mathsf{PG}(2,q^{2})$, where $q$ is a power of the prime $p$.

Lemma 3.2**.**

Let $p$ be a prime.

1. (i)

Let $g$ be an element of the affine linear group $\mathsf{AGL}(1,p^{f})$ such that $o(g)\mid p^{f}-1$. Then $g$ has a unique fixed point $v\in\mathbb{F}_{p^{f}}$ and permutes $\mathbb{F}_{p^{f}}$ in orbits of length $o(g)$. 2. (ii)

Let $S$ be a subgroup of $\mathsf{AGL}(1,p^{f})$ such that $p\nmid|S|$. Then, $S$ is cyclic and $|S|$ divides $p^{f}-1$. Moreover, $S$ has a unique fixed point in $\mathbb{F}_{p^{f}}$. ∎

Lemma 3.3**.**

Let $\ell_{1},\ell_{2}$ be two lines of $\mathsf{PG}(2,q^{2})$ and $P,Q$ be two points off $\ell_{1}\cup\ell_{2}$. Write $Z=\ell_{1}\cap\ell_{2}$ and $V_{i}=\ell_{i}\cap PQ$, $i=1,2$. The perspectivity $\pi_{\ell_{1},P,\ell_{2}}\pi_{\ell_{2},Q,\ell_{1}}$ fixes $Z$ and $V_{1}$ and permutes $\ell_{1}\setminus\{Z,V_{1}\}$ in orbits of equal lengths.

Proof.

Elementary. ∎

Let $S$ be any set of $n+1$ points in the projective plane $\Pi$ of order $n$. A nucleus of $S$ is a point $P$ such that each line of $\Pi$ through $P$ intersects $S$ in a unique point. It follows that $P\not\in S$. We denote by $\mathcal{N}(S)$ the set of all nuclei of $S$.

Let $U=\left(X,B\right)$ be a unital of order $q$ embedded in $\mathsf{PG}{\left(2,q^{2}\right)}$ and let $b_{1},b_{2}\in B$ be two (not necessarily disjoint) blocks of $U$. Denote the lines containing the blocks $b_{1}$ and $b_{2}$ by $\ell_{1}$ and $\ell_{2}$ respectively. Using the notations in [10] let $\mathcal{B}=b_{1}\cup\left(\ell_{2}\setminus b_{2}\right)$: the set $\mathcal{B}$ consists of $q^{2}+1$ non collinear points, it is contained in the union of the lines $\ell_{1}$ and $\ell_{2}$. Note that $Z=\ell_{1}\cap\ell_{2}$ belongs to $\mathcal{B}$. Let $\mathcal{N}{\left(\mathcal{B}\right)}$ denote the set of all nuclei of $\mathcal{B}$. Clearly, if $P$ is a full point w.r.t. to the blocks $b_{1},b_{2}$ then $P$ is a nucleus of $\mathcal{B}$, hence $F_{U}{\left(b_{1},b_{2}\right)}\subseteq\mathcal{N}{\left(\mathcal{B}\right)}$.

The next lemma formulates [10, Propositions 2 and 3] in our setting.

Lemma 3.4**.**

Let $U=\left(X,B\right)$ be a unital of order $q$ embedded in $\mathsf{PG}{\left(2,q^{2}\right)}$ and let $b_{1},b_{2}\in B$ be two blocks of $U$. Denote the lines containing the blocks $b_{1}$ and $b_{2}$ by $\ell_{1}$ and $\ell_{2}$ respectively. Write $Z=\ell_{1}\cap\ell_{2}$ and $\mathcal{B}=b_{1}\cup\left(\ell_{2}\setminus b_{2}\right)$. Define the set $\Gamma_{1}=\left\{\pi_{\ell_{1},P,\ell_{2}}\pi_{\ell_{2},Q,\ell_{1}}\mid P,Q\in\mathcal{N}{\left(\mathcal{B}\right)}\right\}$ where $\mathcal{N}{\left(\mathcal{B}\right)}$ denotes the set of all nuclei of $\mathcal{B}$. Then the following hold:

1. (i)

$\Gamma_{1}$* leaves $b_{1}$ invariant.* 2. (ii)

$\Gamma_{1}$* is a group of affinities of the affine line $\ell_{1}\setminus\left\{Z\right\}$. ∎*

Define the integer $r$ by $q^{2}=p^{r}$. The order of the group $\Gamma_{1}$ is $tp^{h}$, where $p\nmid t$, and $\Gamma_{1}$ is isomorphic to some group $\Gamma=\mathbf{A}\mathbf{B}$ of affinities where $\mathbf{B}$ is an additive subgroup of order $p^{h}$ of $\mathsf{GF}{\left(q^{2}\right)}$ and $\mathbf{A}$ is a multiplicative subgroup of order $t$ of $\mathsf{GF}{\left(q^{2}\right)}$ such that $t\mid p^{\gcd\left(r,h\right)}-1$. Let $m=\left(p^{r-h}-1\right)/t$ and let $\mathbf{B}_{1}\cup\mathbf{O}_{1}\cup\ldots\cup\mathbf{O}_{m}$ be the partition of $\ell_{1}\setminus\left\{Z\right\}$ into $\Gamma_{1}$-orbits. We have by [10, Section 2] that $\mathbf{B}_{1}$ has length $p^{h}$ and for each $i=1,2,\ldots,m$ the orbit $\mathbf{O}_{i}$ has length $tp^{h}$.

Let $\mathcal{B}_{i}=\ell_{i}\cap\mathcal{B}$ for $i=1,2$ and let $\widehat{\mathcal{B}}_{1}=\mathcal{B}_{1}\setminus\left\{Z\right\}$, then $\widehat{\mathcal{B}}_{1}$ is the union of $\Gamma_{1}$-orbits. It follows that the size of $\widehat{\mathcal{B}}_{1}$ must be divisible by $p^{h}$, and we must distinguish between two cases:

1. (1)

If the blocks $b_{1}$ and $b_{2}$ are disjoint, it means $b_{1}=\mathcal{B}_{1}=\widehat{\mathcal{B}}_{1}$, hence $p^{h}\mid q+1$. It is possible only for $h=0$, thus the group $\mathbf{B}$ is trivial. 2. (2)

Otherwise $b_{1}\cap b_{2}=\left\{Z\right\}$, meaning $b_{1}=\mathcal{B}_{1}=\widehat{\mathcal{B}}_{1}\cup\left\{Z\right\}$, hence the size of $\widehat{\mathcal{B}}_{1}$ is $q$. In this case $q=ap^{h}+btp^{h}$, where $b\in\left\{0,1,\ldots,m\right\}$ and $a=1$ or [math], depending on whether $\mathbf{B}_{1}\subseteq\widehat{\mathcal{B}}_{1}$ or not. If $a=0$, then $q=btp^{h}$, and as $p\nmid t$ we have $t=1$, therefore the group $\mathbf{A}$ is trivial.

Lemma 3.5**.**

Let $U=\left(X,B\right)$ be a unital of order $q$ embedded in $\mathsf{PG}{\left(2,q^{2}\right)}$ and let $b_{1},b_{2}\in B$ be two disjoint blocks of $U$. Denote the lines containing the blocks $b_{1}$ and $b_{2}$ by $\ell_{1}$ and $\ell_{2}$ respectively. Write $Z=\ell_{1}\cap\ell_{2}$ and $\mathcal{B}=b_{1}\cup\left(\ell_{2}\setminus b_{2}\right)$. Define the group $\Gamma_{1}$ generated by the perspectivities $\pi_{\ell_{1},P,\ell_{2}}\pi_{\ell_{2},Q,\ell_{1}}$ with $P,Q\in\mathcal{N}{\left(\mathcal{B}\right)}$ where $\mathcal{N}{\left(\mathcal{B}\right)}$ denotes the set of all nuclei of $\mathcal{B}$. Then the following hold:

1. (i)

$p\nmid\left|\Gamma_{1}\right|$. 2. (ii)

$\Gamma_{1}$* is cyclic and $|\Gamma_{1}|\mid q^{2}-1$.* 3. (iii)

$\Gamma_{1}$* has a unique fixed point $V_{1}\not\in b_{1}\cup\left\{Z\right\}$.* 4. (iv)

The set of full points $F_{U}{\left(b_{1},b_{2}\right)}$ is contained in a line $m$ through $V_{1}$ but $Z$.

Proof.

Assume that $\Gamma_{1}$ has an element $\gamma$ of order $p$. Since $b_{1}$ is $\Gamma_{1}$-invariant, $\gamma$ has a fixed point in $b_{1}$, different from $Z$ as $Z\not\in b_{1}$. However, affinities with two fixed points have order dividing $q^{2}-1$. This proves (i).

Together with Lemmas 3.2 and 3.3, (i) implies (ii) and (iii). Notice that Lemma 3.2(i) is needed to show that $V_{1}\not\in b_{1}$.

Since $\mathbf{B}$ is trivial, the set of nuclei $\mathcal{N}{\left(\mathcal{B}\right)}$ is contained in a line $m$ such that $Z\not\in m$ (cf. [10, p. 67]). In particular $F_{U}{\left(b_{1},b_{2}\right)}$ is contained in $m$ as $F_{U}{\left(b_{1},b_{2}\right)}\subseteq\mathcal{N}{\left(\mathcal{B}\right)}$. Furthermore, by Lemma 3.3, for any $P,Q\in\mathcal{N}{\left(\mathcal{B}\right)}$ the line $PQ$ contains $V_{1}$, hence $V_{1}\in m$. This proves (iv). ∎

We can now state and prove the main theorem of this section.

Theorem 3.6**.**

If the unital $U$ of order $q$ is embedded in $\mathsf{PG}{\left(2,q^{2}\right)}$ then it is strongly full point regular.

Proof.

Let us assume that $U$ is embedded in $\mathsf{PG}(2,q^{2})$. Let $b_{1},b_{2}$ be two disjoint blocks of $U$. If $|F_{U}(b_{1},b_{2})|\leq 1$ then we have nothing to prove. Otherwise, by Lemma 3.5 $F_{U}(b_{1},b_{2})$ is contained in a block $c$, which is disjoint to $b_{1}$ and $b_{2}$. Furthermore, $\operatorname{Persp}_{b_{2}}(b_{1})$ is cyclic, its order divides $q^{2}-1$ and $b_{1}$ decomposes into orbits of equal lengths. This means that $(U,b_{1},b_{2})$ is a strongly full point regular triple. ∎

4. Full points of the Hermitian unital

For a prime power $q$, let $\rho$ be a Hermitian polarity of $\mathsf{PG}(2,q^{2})$. Two points $P,Q$ are said to be conjugate if $P\in Q^{\rho}$. Similarly, the lines $\ell,m$ are conjugate if $\ell^{\rho}\in m$. Let $R^{+}$ be the set of pairs $(\ell,m)$, where $\ell,m$ are conjugate lines to each other but not self-conjugate. The projective unitary group $\mathsf{PGU}(3,q)$ acts transitively on $R^{+}$. Given two conjugate lines $\ell_{1},\ell_{2}$, one constructs $\ell_{3}=(\ell_{1}\cap\ell_{2})^{\rho}$, conjugate to both $\ell_{1}$ and $\ell_{2}$. We say that $\ell_{1},\ell_{2},\ell_{3}$ form a polar triangle. The projective unitary group $\mathsf{PGU}(3,q)$ acts transitively on the set of polar triangles. Consider the set ${X}$ of self-conjugate points of $\rho$; $|{X}|=q^{3}+1$. The line $\ell$ intersects ${X}$ in $1$ or $q+1$ points, depending on if $\ell$ is self-conjugate or not. Let $\ell$ be a non self-conjugate line and $m$ be a line connecting $\ell^{\rho}$ and a point $P\in X\cap\ell$. Since $\ell^{\rho}\in P^{\rho}$, we have $m=P^{\rho}$ which must be a self-conjugate line. This means that $(\ell,\ell^{\prime})\in R^{+}$ implies that $\ell\cap\ell^{\prime}\not\in X$. It follows that any non self-conjugate line $\ell$ is contained in exactly $q(q-1)/2$ polar triangles. For further details and background, see [8, Section 7.3]

The abstract Hermitian unital $\mathcal{H}(q)$ is constructed from the set ${X}$ of self-conjugate points of $\rho$. The subsets cut out by the $(q+1)$-secants (not self-conjugate lines) form the set $B$ of blocks of $\mathcal{H}(q)$. Notice that we consider $\mathcal{H}(q)$ as an abstract unital, having a natural embedding in $\mathsf{PG}(2,q^{2})$. The following proposition gives a characterization of the conjugate relation in terms of the abstract unital $\mathcal{H}(q)$ for $q$ even.

Proposition 4.1**.**

Let $q$ be even, let $\rho$ be a Hermitian polarity of $\mathsf{PG}(2,q^{2})$ and let $X$ be the set of self-conjugate points of $\rho$. Let $\ell_{1},\ell_{2}$ be not self-conjugate lines and define the blocks $b_{i}=\ell_{i}\cap X$ of $\mathcal{H}(q)$, $i=1,2$. Then the following hold:

1. (i)

If $\ell_{1},\ell_{2}$ are conjugate, then $F_{\mathcal{H}(q)}(b_{1},b_{2})=b_{3}$, where $b_{3}=\ell_{3}\cap X$ with $\ell_{3}=(\ell_{1}\cap\ell_{2})^{\rho}$. In other words, the blocks contained in a polar triangle form an embedded dual $3$-net of $\mathcal{H}(q)$. 2. (ii)

If $\ell_{1},\ell_{2}$ are not conjugate then either $b_{1}\cap b_{2}\neq\emptyset$, or $|F_{\mathcal{H}(q)}(b_{1},b_{2})|=1$.

Proof.

(i) Up to projective equivalence, we can assume that the matrix of $\rho$ is the identity. Since the unitary group $\mathsf{PGU}(3,q)$ acts transitively on $R^{+}$, we can assume $\ell_{1}:X_{1}=0$ and $\ell_{2}:X_{2}=0$. Then, $\ell_{1}\cap\ell_{2}=(0,0,1)$ and $\ell_{3}:X_{3}=0$. Let $\varepsilon$ be a $(q+1)$th root of unity in $\mathbb{F}_{q^{2}}$. The elements of $b_{s}=\ell_{s}\cap X$, $s=1,2,3$, have the form

[TABLE]

respectively, with $i,j,k=0,1,\ldots,q$. Since the points $A_{i},B_{j},C_{k}$ are collinear if and only if $\varepsilon^{i+j+k}=1$, we see that $A_{i}$ projects from $C_{k}$ to $B_{-i-k}$. In particular, $b_{3}\subseteq F_{\mathcal{H}(q)}(b_{1},b_{2})$, and equality holds by Theorem 3.6.

(ii) The case when $\ell_{1},\ell_{2}$ are not conjugate and $b_{1}\cap b_{2}=\emptyset$ was elaborated in [12, Section 2.2]. ∎

Remark 4.2**.**

Proposition 4.1 shows that for $q$ even, $\mathcal{H}(q)$ has embedded dual $3$-nets. More precisely, any block of $\mathcal{H}(q)$ is contained in $q(q-1)/2$ polar triangles. The group of automorphisms of $\mathcal{H}(q)$ acts transitively on the set of embedded dual $3$-nets.

Let $\rho_{0}$ be a Hermitian polarity of the projective line $\mathsf{PG}(1,q^{2})$. The set of self-conjugate points of $\rho_{0}$ forms a subline $\mathsf{PG}(1,q)$, cf. [8, Lemma 6.2]. Let $\ell$ be a line of $\mathsf{PG}(2,q^{2})$. A Baer subline of $\ell$ is subset of size $q+1$, consisting of self-conjugate points of some Hermitian polarity $\rho$ of $\mathsf{PG}(2,q^{2})$. Equivalently, a Baer subline $S$ is isomorphic to $\mathsf{PG}(1,q)$, and $S=\ell\cap\Pi$ for some line $\ell$ and a Baer subplane $\Pi$.

Proposition 4.3**.**

Let $U=(X,B)$ be an abstract unital of order $q$, embedded in $\mathsf{PG}(2,q^{2})$. Let $b_{1},b_{2},b_{3}$ form an embedded dual $3$-net. Then $b_{1},b_{2},b_{3}$ are Baer sublines.

Proof.

Let $\ell$ be the projective line containing $b_{1}$. By Theorem 3.6, $C=\operatorname{Persp}_{b_{2}}(b_{1})$ is a cyclic subgroup of order $q+1$, preserving $b_{1}$. Since $C$ is obtained using projections in $\mathsf{PG}(2,q^{2})$, it is a subgroup of the projectivity group of $\ell$. By the arguments of [12, Section 3] one shows that $b_{1}$ is a Baer subline of $\ell$. ∎

Remark 4.4**.**

Let $q$ be even, and consider an arbitrary embedding of the Hermitian unital $\mathcal{H}(q)$ in $\mathsf{PG}(2,q^{2})$. By Remark 4.2 and Proposition 4.3, all blocks correspond to Baer sublines of $\mathsf{PG}(2,q^{2})$. Using the characterization of Hermitian curves from [7, 15], this observation gives a simple proof of the main theorem in [12] in the even $q$ case.

5. Full points and dual $3$-nets of known small unitals

In this section we present computational results on the structure of full points of known small unitals. More precisely, we study the following classes of abstract unitals of order at most $6$:

**Class BBT: **

909 unitals of order 3 by Betten, Betten and Tonchev [6],

**Class KRC: **

4466 unitals of order 3 with nontrivial automorphism groups by Krčadinac [13],

**Class KNP: **

1777 unitals of order 4 by Krčadinac, Nakić and Pavčević [14],

**Class BB: **

two cyclic unitals of order 4 and 6 by Bagchi and Bagchi [1].

Notice that KRC contains all abstract unitals of order $3$ with a nontrivial automorphism group. As mentioned in [13], 722 of the BBT unitals appear in KRC. Moreover, the cyclic BB unital of order $4$ is contained in KNP. The BB unital of order $6$ has no full points, therefore we omit the BB class from the tables of this section. We access the libraries of small unitals and carry out the computations using the GAP4 package UnitalSZ [16].

5.1. The number of full points and the structure of the group of perspectivities

We only consider disjoint pairs of blocks admitting at least two full points as for only one full point the perspectivitiy group is trivial. In Tables 1, 2 and 3 we summarize the existing number of full points, the structure of the group of perspectivities and the number of unitals with such pairs for each library (BBT, KRC, KNP).

5.2. The structure of the full points

The structure of the full points is only interesting when there is at least 3 of them, hence the BBT unitals are out of our scope. Even the case of 3 full points is simple: they are either contained in a block or not. As KRC unitals admit at most 3 full points, we are only interested in the KNP unitals.

The computation in [16] showed that if there are 4 or 5 full points (in the case of disjoint blocks) then either the whole set of full points is contained in a single block, or no three points are collinear. Similarly in the case of 6 full points either 5 of the full points form a block or no 3 of them are collinear. Now by “collinear” we mean that the points form a subset of some block of the unital.

5.3. Unitals with large full point sets

Let us denote by $\Omega$ the subset of unitals with at least one large full point set, that is, $|F_{U}(b_{1},b_{2})|\geq 3$ for a pair $(b_{1},b_{2})$ of disjoint blocks. We have seen that $\Omega$ is the empty set for BBT unitals. By Table 2, $|\Omega|=6$ for KRC unitals. Hence, the interesting case is the KNP library, where the size of $\Omega$ is 206. In Table 4 we present the number of KNP unitals with some restrictions on the structure of full points. Clearly $A\subseteq B$, $C\subseteq\overline{B}$ and $\Omega=B\cup\overline{B}$.

5.4. Full point regularity

In Table 5 one sees how many of the unitals of the different libraries are full point regular (FPR) and strongly full point regular (SFPR). In fact, if a unital is not strongly full point regular then is not embeddable into $\mathsf{PG}(2,q^{2})$. Hence 94 BBT unitals, 385 KRC unitals and 195 KNP unitals are definitely not embeddable into $\mathsf{PG}(2,q^{2})$. Notice that [2] proves a much stronger result, where the authors show that there are just two orbits of unitals in $\mathsf{PG}(2,16)$, containing the Hermitian unitals and Buekenhout–Metz unitals, respectively.

5.5. Embedded dual 3-nets

By Proposition 2.6(ii), one can find embedded dual 3-nets only among the KNP unitals. The computation shows us that the latin squares constructed from the dual 3-nets are always of cyclic type, namely, any embedded dual 3-net is cyclic in the KNP library. However, we constructed a new unital of order 4 with a non-cyclic embedded dual 3-net, cf. Appendix A.

Appendix A Unital of order $4$ with non-cyclic embedded dual $3$-net

⬇

LoadPackage(”UnitalSZ”);

bls:=[[1,2,55,64,65], [1,3,32,46,63], [1,4,7,34,45], [1,5,11,31,44],

[1,6,12,19,54], [1,8,38,47,50], [1,9,24,27,40], [1,10,20,48,53],

[1,13,17,49,57], [1,14,15,16,29], [1,18,33,43,58], [1,21,23,25,37],

[1,22,51,56,60], [1,26,30,39,52], [1,28,36,41,62], [1,35,42,59,61],

[2,3,6,30,58], [2,4,14,54,60], [2,5,29,46,47], [2,7,13,48,59],

[2,8,34,37,40], [2,9,10,18,31], [2,11,19,32,52], [2,12,20,50,57],

[2,15,21,43,62], [2,16,23,27,28], [2,17,33,45,61], [2,22,24,25,26],

[2,35,38,39,41], [2,36,49,53,56], [2,42,44,51,63], [3,4,19,23,33],

[3,5,10,39,59], [3,7,22,49,52], [3,8,14,48,65], [3,9,25,29,60],

[3,11,15,20,34], [3,12,13,16,61], [3,17,24,28,44], [3,18,47,53,57],

[3,21,36,40,42], [3,26,37,38,43], [3,27,35,56,64], [3,31,45,55,62],

[3,41,50,51,54], [4,5,41,52,53], [4,6,26,31,47], [4,8,16,36,57],

[4,9,56,58,59], [4,10,28,46,65], [4,11,21,50,64], [4,12,35,44,62],

[4,13,30,32,51], [4,15,37,39,61], [4,17,18,20,25], [4,22,27,38,48],

[4,24,42,49,55], [4,29,40,43,63], [5,6,7,32,37], [5,8,42,54,58],

[5,9,12,17,51], [5,13,27,36,63], [5,14,22,61,62], [5,15,25,40,49],

[5,16,19,20,26], [5,18,21,28,38], [5,23,30,55,60], [5,24,33,48,64],

[5,34,35,57,65], [5,43,45,50,56], [6,8,56,62,63], [6,9,21,61,65],

[6,10,14,40,41], [6,11,25,43,51], [6,13,38,44,55], [6,15,42,46,57],

[6,16,22,34,64], [6,17,23,36,52], [6,18,48,49,60], [6,20,28,45,59],

[6,24,35,50,53], [6,27,29,33,39], [7,8,20,24,51], [7,9,41,63,64],

[7,10,11,42,60], [7,12,15,55,56], [7,14,23,26,35], [7,16,44,46,53],

[7,17,29,38,62], [7,18,19,39,50], [7,21,27,31,57], [7,25,47,58,65],

[7,28,30,33,40], [7,36,43,54,61], [8,9,11,13,46], [8,10,12,45,52],

[8,15,18,27,59], [8,17,21,35,60], [8,19,43,49,64], [8,22,29,30,53],

[8,23,32,39,44], [8,25,31,33,41], [8,26,28,55,61], [9,14,52,55,57],

[9,15,19,28,53], [9,16,35,43,47], [9,20,22,36,39], [9,23,48,50,62],

[9,26,32,33,42], [9,30,34,38,54], [9,37,44,45,49], [10,13,23,34,43],

[10,15,17,30,64], [10,16,21,32,56], [10,19,25,35,55], [10,22,33,54,57],

[10,24,36,37,47], [10,26,27,51,62], [10,29,44,50,61], [10,38,49,58,63],

[11,12,33,38,59], [11,14,39,47,56], [11,16,18,54,62], [11,17,22,41,65],

[11,23,24,29,57], [11,26,36,45,48], [11,27,30,49,61], [11,28,35,37,63],

[11,40,53,55,58], [12,14,24,30,43], [12,18,23,42,65], [12,21,26,41,49],

[12,22,28,32,47], [12,25,34,48,63], [12,27,37,53,60], [12,29,31,36,58],

[12,39,40,46,64], [13,14,21,33,53], [13,15,41,45,47], [13,18,26,29,64],

[13,19,24,31,56], [13,20,35,52,58], [13,22,37,42,50], [13,25,28,39,54],

[13,40,60,62,65], [14,17,19,59,63], [14,18,37,46,51], [14,20,31,38,42],

[14,25,36,44,64], [14,27,32,45,58], [14,28,34,49,50], [15,22,23,31,63],

[15,24,32,38,65], [15,26,50,58,60], [15,33,35,36,51], [15,44,48,52,54],

[16,17,37,48,58], [16,24,41,59,60], [16,25,30,42,45], [16,31,39,49,65],

[16,33,50,55,63], [16,38,40,51,52], [17,26,34,46,56], [17,27,47,54,55],

[17,31,32,40,50], [17,39,42,43,53], [18,22,35,40,45], [18,24,52,61,63],

[18,30,41,44,56], [18,32,34,36,55], [19,21,22,44,58], [19,27,34,41,42],

[19,29,45,51,65], [19,30,37,57,62], [19,36,38,46,60], [19,40,47,48,61],

[20,21,30,47,63], [20,23,40,54,56], [20,27,43,44,65], [20,29,37,41,55],

[20,32,60,61,64], [20,33,46,49,62], [21,24,45,46,54], [21,29,34,52,59],

[21,39,48,51,55], [22,43,46,55,59], [23,38,45,53,64], [23,41,46,58,61],

[23,47,49,51,59], [24,34,39,58,62], [25,27,46,50,52], [25,32,53,59,62],

[25,38,56,57,61], [26,40,44,57,59], [26,53,54,63,65], [28,29,42,48,56],

[28,31,43,52,60], [28,51,57,58,64], [29,32,35,49,54], [30,31,35,46,48],

[30,36,50,59,65], [31,34,51,53,61], [31,37,54,59,64], [32,41,43,48,57],

[33,34,44,47,60], [33,37,52,56,65], [39,45,57,60,63], [42,47,52,62,64]

];;

u:=AbstractUnitalByDesignBlocks(bls);

t:=BlocksOfUnital(u){[1,33,200]};

StructureDescription(PerspectivityGroupOfUnitalsBlocks(u,t[1],t[2],t[3]));